Is there any shortcut for this lengthy problem?

[Benjamin_harsh]

Homework Statement

Find the moment of Inertia of the cross-sectional area of an I section about its centroidal axis.

Relevant Equations

$I_{XX1} = I_{.G1.X} + A_{1}.Y^2$

Find the moment of Inertia of the cross-sectional area of an I section about its centroidal axis:

Sol: Here A1, A2 and A3 are the areas:

Centriod ; $X_{C} = \large \frac {A_{1}X_{1} + A_{2}X_{2}+A_{3}X_{3}}{A_{1} + A{2}+ A_{3}}$ $= 15 cm$
Centriod ; $Y_{C} = \large \frac {A_{1}Y_{1} + A_{2}Y_{2}+A_{3}Y_{3}}{A_{1} + A{2}+ A_{3}}$ $= 10.96 cm$

Moment of Inertia w.r.t Centroid X-X:
$I_{XX} = I_{XX1} + I_{XX2} + I_{XX3}$

$I_{XX1} = I_{G1.X} + A_{1}Y^{2}$

$= I_{G1.X} + A_{1}.(Y_{1} - \overline {Y})^{2}$

$= \large \frac {30*5}{12}$ $+ 150(25 - 10.96^{2})$ (Here $I_{G1.X} = \large \frac {b.d^3}{12})$

$= 11048.24 cm^{4}$

$I_{XX2} = I_{G2.X} + A_{2}Y^{2}$

$= I_{G1.X} + A_{2}.(Y_{2} - \overline {Y})^{2}$

$= \large \frac {5*15}{12}$ $+ 75(12.5 - 10.96^{2})$

$=1584.12 cm^{4}$

$I_{XX3} = I_{G3.X} + A_{3}.Y^{2}$

$= I_{G3.X} + A_{3}.(Y_{3} - \overline {Y})^{2}$

$=\large \frac {20*5}{12}$ $+ 100(22.5 - 10.96^{2})$

$= 13525.25 cm^{4}$

$I_{xx} = 11048.28 + 1584.12 + 13525.5 = 26137.86 cm^{4}$

Moment of Inertia w.r.t Centroid Y-Y:

$I_{YY} = I_{YY1} + I_{YY2} + I_{YY3}$

$I_{YY1} = I_{G1 }+ A_{1}.X^{2}$

= $I_{G1.Y} + A_{1}. (X_{1} - \overline X)^{2}$ (Here $I_{G1.X} = \large \frac{d.b^{3}}{12}$)

$= \large \frac {5*30^{3}}{12}$ + $(150)(15 - 15)^{2}$

$= 11250 cm^{4}$

$I_{YY2} = I_{G2} + A_{2}.X^{2}$

= $I_{G2.Y} + A_{2}. (x^{2} - \overline X)^{2}$

= $\large \frac {15*5^{3}}{12}$ + $(75)(15-15)^{2}$

= $156.25 cm^{4}$

$I_{YY3} = I_{G3} + A_{3}.X^{2}$

= $I_{G3.Y} + A_{3}. (x^{2} - \overline X)^{2}$

= $\large \frac {5*20^{3}}{12}$ + $(100)(15-15)^{2}$

= $3333.33 cm^{4}$

$I_{yy} = 11250 + 156.25 + 3333.33 = 14739.58 cm^{4}$

 

[BvU]

What is the axis of rotation ? Where is the centroidal axis located in the drawing ?

How come the moment of inertia has dimension length⁴ ?

 

[Benjamin_harsh]

Where is the centroidal axis located in the drawing ?

X, Y = 15, 10.96

 

[BvU]

OK.
I was distracted by the dimension, but I suppose mass moment of inertia and area moment can be linked through a density.

So you added a bunch of contributions, making good use of the parallel axis theorem.

Is there a question? Or do you want to hire PF to check your math ?

Magnitude seems a bit low to me: upper rect 100 cm² at 12 cm alone is 15000 cm⁴ already. Or is it me now making a counting error ?

Number of digits is definitely too many: you have 1 and 2 digits dimensions, so you can never get 7 digits accuracy in $I$

For clarity you might explain what X_i,Y_i and A_i stand for...

It looks as if they have a different meaning in you relevant equation.

Your relevant equation itself is also incomplete: it does not show what the definition of $I$ is, but just renders the parrallel axis theorem in an unknown notation.

$\$

 

[Benjamin_harsh]

Can I calculate same question with integration method? If so, how?

 

[haruspex]

$= I_{G1.X} + A_{1}.(Y_{1} - \overline {Y})^{2}$
$= \large \frac {30*5}{12}$ $+ 150(25 - 10.96^{2})$ (Here $I_{G1.X} = \large \frac {b.d^3}{12})$

Seem to be a few problems there.
You correctly refer to $\frac {b.d^3}{12}$, but I don't see anything cubed on the left.
Shouldn’t $(25 - 10.96^{2})$ be $(2.5 - 10.96)^{2}$?

 

[Benjamin_harsh]

Seem to be a few problems there.
You correctly refer to $\frac {b.d^3}{12}$, but I don't see anything cubed on the left.
Shouldn’t $(25 - 10.96^{2})$ be $(2.5 - 10.96)^{2}$?

Now I can't edit my post. Please tell me is it possible to solve this problem through integration method or not?

 

[haruspex]

Now I can't edit my post. Please tell me is it possible to solve this problem through integration method or not?

Of course it can be solved by integration, but that is not going to make it any easier.

 

[Benjamin_harsh]

it can be solved by integration, but that is not going to make it any easier.

I googled and but none helped me. Please will you solution using integration.

 

[haruspex]

I googled and but none helped me. Please will you solution using integration.

In post #1 you wrote you were looking for a simpler solution. All that integration will do is be a longer route to the formulas you were using in the first place.
If you have changed your purpose and now just wish to understand how to derive those formulas by integration I will help you, but if you are hoping it will provide an easier solution to the given problem you are wasting your time - and mine.

 

[Benjamin_harsh]

In post #1 you ...mine.

Are you simply telling my solution is easiest way ?

 

[haruspex]

Are you simply telling my solution is easiest way ?

I am telling you that using integration isn't going to make it easier.
But maybe there is an easier way than post #1.

First, you need to clarify the question. As I remarked on another thread, "centroidal axis" means any axis through the centroid. If you mean the polar axis, i.e. the centroidal axis normal to the plane, then why not start with the moments about the polar axes of the individual rectangles instead of dealing with XX and YY separately?

 

[Benjamin_harsh]

Using the parallel axis theorem will make this easier?

 

[haruspex]

Using the parallel axis theorem will make this easier?

You used that in your post #1. You can also use it with the moment about the polar axis.

 

[Benjamin_harsh]

You can also use it with the moment about the polar axis.

Will I get same answer as post #1?

 

[haruspex]

Will I get same answer as post #1?

In principle you should, but in post #6 I pointed out what appear to be errors in your working in post #1, so probably not.

 

[Benjamin_harsh]

Can you tell how $X_{2}$ and $X_{3}$ is $15$?

 

[haruspex]

Can you tell how $X_{2}$ and $X_{3}$ is $15$?

As in the diagram of the other question, horizontal displacements are not shown, so you need to assume all three shapes are centred on the same x coordinate. If the origin is the lower left corner of the bottom rectangle then that coordinate is 15cm.

 

[Benjamin_harsh]

so you need to assume all three shapes are centred on the same x coordinate. If the origin is the lower left corner of the bottom rectangle then that coordinate is 15cm.

What does centred mean? centroid of 3 shapes lie in same point?

 

[BvU]

On the same x value, i.e. a vertical line